Exercise 5, § 6

Let $E$ be a vector space of finite dimension $n$ over a field $K$. (a) Show that for every integer $m$ the symmetric power $S^m(E)$ and the vector space $S'_m(E)$ of symmetric contravariant tensors of order $m$ have the same dimension $\binom{m+n-1}{n-1} = \binom{n+m-1}{m}$. (b) If $K$ is of characteristic $p > 0$, show that the vector space $S''_p(E)$ of symmetrized tensors of order $p$ has dimension $\binom{n}{p}$ (observe that if, in a tensor product $x_1 \otimes x_2 \otimes \cdots \otimes x_p$ of $p$ vectors of $E$, $x_i = x_j$ for an ordered pair of distinct indices $i, j$, then the symmetrization of this product is zero; use the fact that if $(H_i)_{1 \leq i \leq r}$ is a partition of $\{1, 2, \ldots, p\}$ into non-empty sets, comprising at least two sets, the subgroup of $S_p$ leaving each of the $H_i$ stable has order not divisible by $p$). On the other hand, the canonical image in $S^p(E)$ of the space $S'_p(E)$ of symmetric tensors of order $p$ is of dimension $n$ and is generated by the images of the tensors $x \otimes x \otimes \cdots \otimes x$ ($p$ times), where $x \in E$ (use the same remark); the canonical mapping of $S'_p(E)$ into $S^p(E)$ is therefore not bijective although the two vector spaces have the same dimension.